I can not figure out how to solve the following problem :
$ |x^2 +x| + |3x-9| = x+13 $
should I solve it through factorization ?
$\endgroup$ 43 Answers
$\begingroup$hint: write the equation in the form $$|x||x+1|+3|x-3|=x+13$$ and do case work. the solution is given by $$x=-1$$ or $$x=\frac{1}{2}\left(\sqrt{97}-3\right)$$
$\endgroup$ 1 $\begingroup$Four cases:
1)$x^2+x\ge 0$ and $3x-9\ge 0$ this means $x\ge 3$
If $x \ge 3$ then
$|x^2+x|+|3x-9|=x+13$
$x^2+x+3x-9=x+13$
$x^2+3x-22=0$
$x=\frac {-3\pm \sqrt {9+88}}2$ but as $x \ge 3$
$x=\frac {\sqrt {97}-3} 2$
2) $x^2+x \ge 0 $ and $3x-9 <0$. That means $0\le x <3$ or $x \le -1$.
If $0\le x <3$ or $x \le -1$ then
$|x^2+x|+|3x-9|=x+13$
$x^2+x -3x+9=x+13$
$x^2-3x-4=0$
$(x-4)(x+1)=0$ and as $x <3$, $x=-1$.
3) $x^2 +x <0$ and $3x-9\ge 0$. That means $-1<x <0 $ and $x\ge 3$. Nonsense.
4)$x^2+x <0 $ and $3x-9 < 0$. That means $-1 <x <0$.
Then $|x^2+x |+|3x-9|=x+13$
$-x-x^2-3x+9=x+13$
$x^2+5x+22=0$
So $\frac {-5\pm \sqrt{25 -88}}2$ which is not a real number.
$\endgroup$ $\begingroup$Hint -
Case 1 -
When $|3x - 9| = 3x - 9$
$x^2 + x + 3x - 9 = x + 13$
$x^2 + 3x - 22 = 0$
Now factorise.
Case 2 -
When $|3x - 9| = - 3x + 9$
$x^2 + x - 3x + 9 = x + 13$
$x^2 - 3x - 4 = 0$
Now factorise.
Similarly cases for $|x^2 + x|$ when it is equal to $-x^2 - x$.
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